Prove that if d = gcd(m,n) then $\phi(mn)=\phi(m)*\phi(n)/d$

Question

So if m and n are relatively prime, then the $\phi(mn)=\phi(m)*\phi(n)$ but what happens when $d > 1$?

Answer 1

Denote $P$ as the product of primes common to $m$ and $n$. Then $\phi(mn) = P \phi(m) \phi(n)/\phi(P)$ which generalizes all of this. Found in Niven's book and easy to prove by writing out the factorization of $m,n$.